I'd strongly agree with that, mainly because, while points with this property exist, they are not necessarily unique. The non-uniqueness is a big issue.

It was a typo! And it has been fixed.

It is because beach/beach is the surplus-maximizing result. Any Pareto-optimal bargaining solution where money is involved will involve the surplus-maximizing result being played, and a side payment occuring.

I have a reduction of this problem to a (hopefully) simpler problem. First up, establish the notation used.

[n] refers to the set $\left\{\mathrm{1...}n\right\}$. $n$ is the number of candidates. Use $C$ as an abbreviation for the space $\Delta \left[n\right]$, it's the space of probability distributions over the candidates. View $C$ as embedded in $R^{n-1}$, and set the origin at the center of $C$.

At this point, we can note that we can biject the following: 
1: Functions of type $\left[n\right]\to [0,1]$ 
2: Affine functions of type $C\to [0,1]$ 
3: Functions of the form $\lambda x.⟨a,x⟩+c$, where $x,a\in R^{n-1}$, and and $c\in R$, and everything's suitably set so that these functions are bounded in $[0,1]$ over $C$. (basically, we extend our affine function to the entire space with the Hahn-Banach theorem, and use that every affine function can be written as a linear function plus a constant) We can reexpress our distribution $V$ over utility functions as a distribution over these normal vectors $a$.

Now, we can reexpress the conjecture as follows. Is it the case that there exists a $\mu :\Delta C$ s.t. for all $\nu :\Delta C$, we have 

 Where $\text{sgn}$ is the function that's -1 if the quantity is negative, 0 if 0, and 1 if the quantity is positive. To see the equivalence to the original formulation, we can rewrite things as 

 Where the bold 1 is an indicator function. And split up the expectation and realize that this is a probability, so we get 

 And this then rephrases as 

 Which was the original formulation of the problem.

Abbreviating the function $E_{x,y,a\sim \mu \times \nu \times V}\left[\text{sgn}\right(⟨a,x-y⟩\left)\right]$ as $f(\mu ,\nu )$, then a necessary condition to have a $\mu :\Delta C$ that dominates everything is that 

 If you have this property, then you might not necessarily have an optimal $\mu$ that dominates everything, but there are $\mu$ that get a worst-case expectation arbitrarily close to 0. Namely, even if the worst possible $\nu$ is selected, then the violation of the defining domination inequality happens with arbitrarily small magnitude. There might not be an optimal lottery-lottery, but there are lottery-lotteries arbitrarily close to optimal where this closeness-to-optimality is uniform over every foe. Which seems good enough to me. So I'll be focused on proving this slightly easier statement and glossing over the subtle distinction between that, and the existence of truly optimal lottery-lotteries.

As it turns out, this slightly easier statement (that sup inf is 0 or higher) can be outright proven assuming the following conjecture.

Stably-Good-Response Conjecture: For every $\nu :\Delta C$, and $ϵ>0$, there exists a $\mu :\Delta C$ and a $\delta >0$ s.t. 

Pretty much, for any desired level of suckage and any foe $\nu$, there's a probability distribution $\mu$ you can pick which isn't just a good response (this always exists, just pick $\nu$ itself), but a stably good response, in the sense that there's some nonzero level of perturbation to the foe where $\mu$ remains a good response no matter how the foe is perturbed.

Theorem 1 Assuming the Stably-Good-Response Conjecture, ${sup}_{\mu }{inf}_{\nu }f(\mu ,\nu )\ge 0$.

I'll derive a contradiction from the assumption that $0>{sup}_{\mu }{inf}_{\nu }f(\mu ,\nu )$. Accordingly, assume the strict inequality.

In such a case, there is some $ϵ$ s.t. $0>-ϵ>{sup}_{\mu }{inf}_{\nu }f(\mu ,\nu )$. Let the set $A_{\mu }:=\left\{\nu |f\right(\mu ,\nu )>-ϵ\}$. Now, every $\nu$ lies in the interior of $A_{\mu }$ for some $\mu$, by the Stably-Good-Response Conjecture. Since $\Delta C$ is a compact set, we can isolate a finite subcover and get some finite set $M$ of probability distributions $\mu$ s.t. $\forall \nu \exists \mu \in M:f(\mu ,\nu )>-ϵ$.

Now, let the set $B_{\nu }:=\{\mu \in c.h(M\left)|f\right(\mu ,\nu )<-ϵ\}$. Since $-ϵ>{sup}_{\mu }{inf}_{\nu }f(\mu ,\nu )$, this family of sets manages to cover all of $c.h\left(M\right)$ (convex hull of our finite set.) Further, for any fixed $\nu$, $f(\mu ,\nu )$ is a continuous function $c.h\left(M\right)\to R$ (a bit nonobvious, but true nontheless because there's only finitely many vertices to worry about). Due to continuity, all the sets $B_{\nu }$ will be open. Since we have an open cover of $c.h\left(M\right)$, which is a finite simplex (and thus compact), we can isolate a finite subcover, to get a finite set $N$ of $\nu$ s.t. $\forall \mu \in c.h\left(M\right)\exists \nu \in N:f(\mu ,\nu )<-ϵ$. And now we can go 

 The first strict inequality was from how all $\mu \in c.h\left(M\right)$ had some $\nu \in N$ which made $f(\mu ,\nu )$ get a bad score. The $\ge$ was from expanding the set of options. The $=$ was from how $f$ is a continuous linear function when restricted to $c.h\left(M\right)\times c.h\left(N\right)$, both of which are compact convex sets, so the minimax theorem can be applied. Then the next $\ge$ was from restricting the set of options, and the $>$ was from how every $\nu \in \Delta C$ had some $\mu \in M$ that'd make $f(\mu ,\nu )$ get a good score, by construction of $M$ (and compactness to make the inequality a strict one).

But wait, we just showed $-ϵ>-ϵ$, that's a contradiction. Therefore, our original assumption must have been wrong. Said original assumption was that $0>{sup}_{\mu }{inf}_{\nu }f(\mu ,\nu )$, so negating it, we've proved that 

 As desired.

With this.

What I mean is, the players hit each other up, are like "yo, let's decide on which ROSE point in the object-level game we're heading towards"
Of course, they don't manage to settle on what equilibrium to go for in the resulting bargaining game, because, again, multiple ROSE points might show up in the bargaining game. 
But, the ROSE points in the bargaining game are in a restricted enough zone (what with that whole "must be better than the random-dictator point" thing) to seriously constrain the possibilities in the object-level game. "Worst ROSE point for Alice in the object-level-game" is a whole lot worse for her than "Worst ROSE point for Alice in the bargaining game about what to do in the object-level-game".

So, the players should be able to ratchet up their disagreement point and go "even if the next round of bargaining fails, at least we can promise that everyone does this well, right? Sure, everyone's going for their own idea of fairness, but even if Alice ends up with her worst ROSE point in the bargaining game, her utility is going to be at least this high, and similar for everyone else."

And so, each step of bargaining that successfully happens ratchets up the disagreement point closer to the Pareto frontier, in a way that should quickly converge. If someone disagrees on step 3, then the step-3 disagreement point gets played, which isn't that short of the Pareto frontier. And if someone doesn't have time for all this bargaining, they can just break things off at step 10 or something, that's just about as good as going all the way to infinity.

Or at least, it should work like this. I haven't proved that it does, and it depends on things like "what does ROSE bargaining look like for n players" and "does the random-dictator-point-dominance thing still hold in the n-player case" and "what's the analogue of that strategy where you block your foe from getting more than X utility, when there are multiple foes?". But this disagreement-point ratcheting is a strategy that address your worries with "ever-smaller pieces of the problem live on higher meta-levels, so the process of adding layers of meta actually converges to solving the problem, and breaking it off early solves most of the problem"

Regarding your last comment, yes, you could always just have a foe that's a jerk, but you can at least act so they don't gain from being an jerk, in a way robust against you and a foe having slightly different definitions of "jerk".